Question

A trolley is being pulled up an inclined plane by a man sitting on it (as shown in the figure). He applies a force of $250N$. If the combined mass of the man and trolley is $100kg$, the acceleration of the trolley will be $\left[\sin {15}^{\circ }=0.26\right]$ :

• A. $2.4\text{m/}{\text{s}}^{\text{2}}$
• B. $\text{9}\text{.4 m/}{\text{s}}^{\text{2}}$
• C. $\text{6}\text{.9 m/}{\text{s}}^{\text{2}}$
• D. $\text{4}\text{.9 m/}{\text{s}}^{\text{2}}$

Answer 1

Answer: (D)

$\text{4}\text{.9 m/}{\text{s}}^{\text{2}}$

Force applied by man rope is given by $F=250N$

Net force applied by tension force on trolley system,

$F_{n}et=3T=3\ast 250=750$..................(1)

Total weight along the inclined plane downwards,

$W=mgsin\theta$

=$100\ast 9.8\ast sin{15}^0=253.6N$................(2)

Net upward force along the incline,

$F_{net}=3T-W=750-253.6=496.35N$, (from(1) and (2))

Now, Newton's second law,

F=ma

$496.5=100\ast a$

solving, $a=4.96m/s^2$